What is the sum from n=1 to infinity}\sqrt{n of-1 ?

The sum from n=1 to infinity}\sqrt{n of-1 is diverges

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sum from n=1 to infinity}\sqrt{n of-1

n = 1 \sum \infty n - 1

diverges

Solution steps

n = 1 \sum \infty n - 1

Apply Series Divergence Test:

diverges

= diverges

\frac{d}{d x} ((sin (x))^{6})

\frac{d}{d x} (\frac{x ^{2}}{y})

\int \frac{1}{1 - u ^{2}} d u

\frac{x - x ^{2}}{x} - 2 cot (x)

What is the sum from n=1 to infinity}\sqrt{n of-1 ?
The sum from n=1 to infinity}\sqrt{n of-1 is diverges